We start with the upper bound. Suppose among all partitions of size \(n\)  that \(\lambda \)  maximizes \(\widehat{N}(\lambda )\) . Then it suffices to show that \(\ell (\lambda )=n\) , since in this case \(\lambda = \left\langle 1^n \right\rangle \)  and \(\widehat{N}(\lambda )=2^n\) . Suppose then that \(\ell (\lambda )=k<n\) . There must be a part of \(\lambda \)  that is larger than one, so let \(r\)  denote a largest such part. We can then inductively assume that the partition is of the form \(\lambda =\left<1^{n-r}, r\right>\)  (for \(n=1\) , \(n=2\) , etc. the hypothesis is easily verified). Now we have that \(\widehat{N}(\left<1^{n-r}, r\right>)=2^{n-r}p_r\) , but a simple application of Bertrand's postulate shows that \(p_r<2 p_{r-1}\) . This shows that as long as \(k<n\)  we can increase \(k\)  while only increasing the integer our partition is mapped to by \(\widehat{N}\) ; in other words, we must have \(k=n\) .
For the lower bound, we argue similarly except now assuming that \(\lambda \)  minimizes \(\widehat{N}(\lambda )\)  among all partitions of size \(n\) . It suffices to show that \(\ell (\lambda )=1\) , and we notice that for \(n=1\) , \(n=2\) , \(n=3\) , etc. this is the case. Thus we can inductively assume that \(\widehat{N}(\lambda )=p_rp_{n-r}\) . It now suffices to show that \(p_a p_b > p_{a+b}\)  for all positive integers \(a\)  and \(b\) . Notice that if \(a=1\)  this is simply Bertrand's postulate, so assume without loss of generality that \(1<a\le b\) . Thus \(p_{a+b} \le p_{2b}\)  and \(p_a p_b \ge 3p_b\) . We now seek to show \(p_{2b} \le 3 p_b\)  for all \(b\) . By a result of Rosser and Shoenfeld [1]} we have for \(n\ge 2\)
\( p_n > n (\log n + \log \log n -3/2)\) 
and for \(n\ge 20\)
\(p_n < n ( \log n + \log \log n - 1/2).\) 
A simple computation shows that \(p_{2b} \le 3p_{b}\)  for all \(b\ge 48\) . The remaining cases can be quickly verified with a computer.
